Can someone help me with this, I don't really understand how to do it, if you help me with one step by step i could figure out the rest.
For each real function described , give all the values of v for which the function has singularities and tell whether the singularities are removable. If the singularities are removable, redefine the function to make it continuous.
This a singularity when cos(x) = 0. cos(x)=0 when x is an odd integer multiple of pi.
That is when x=((2n+1)pi)/2, n is any integer.
This singularity is not removable.
The denominator is 0 when 3a^2 - 2a - 8 = 0
factor this: 3a^2 - 2a - 8 = 0
3a^2 -6a + 4a - 8 = 0 3a(a - 2) + 4(a - 2) = 0
(3a + 4)(a - 2) = 0
a = -4/3 a = 2.
So h(a) has singularities at a = -4/3 a = 2.
The singularities are not removable. Another thing is that if the denominator is not 0
at the singularities then it's definitely not removable.
d)j(y)=(y-1)(y+2)(y-3) it has no singularity since it's a polynomial and polynomial
are continuous for all real numbers